CHM 234: Worksheet #1
Due: Tuesday, August 30th in class
Fall 2011
Dr. Halligan
A. Line Angle/Skeletal Structures
One of the most important things to become comfortable with right away in Organic
Chemistry is line angle structures (also called “skeletal structures”). In a line angle
structure, the carbons and hydrogens are not drawn in. It is assumed that each carbon
has a sufficient number of hydrogens so that it has a total of 4 bonds. Since hydrogen
only has 1 valence electron, it can only have one bond. Always include lone pair
electrons for the heteroatoms (i.e. N, O, halogens, etc.). This will be very important
for writing reaction mechanisms. Geometry is important when you draw line angle
structures so take note of that in the examples provided.
Examples:
a. Here are several ways to represent 2-Methylpentane. The structure on the left is the
line angle (skeletal) structure. They are a lot quicker to draw and easier to work with
for this course.
H
H
H
C
H H
H
H
C
C
H
C
C
C
H HH HH H
2-Methylpentane
CH3CH2CH2CH(CH3)2
C6H14
b. Here are some examples of structures that include heteroatoms. Do you recognize
some of these functional groups? (i.e. ketone, amide, alcohol, alkene, alkyne, ether,
ester, amine, nitrile, aromatic ring, etc.) Look in chapter 3 for assistance.
H2 O
C
C
H2C CH NH2
H2C CH2
O
1.
NH2
H HO H
H
C
C
H
C
C
C
H H HH H
OH
2.
O
3.
O
C6H11NO
C5H10O
H H
H H
C
O
C
H H
C
C
C
H
H
C
C
C
C
O H H
H HH H
C9H14O2
Problems:
1. Determine the molecular formula for the following structures (a-f). Write in the
missing lone pairs of electrons.
B. Degree of Unsaturation
When determining a possible structure for a molecular formula, it is helpful to
calculate the total number of rings + pi bonds (Degree of Unsaturation) that the
structure will have.
Example:
Provide 3 possible skeletal structures for C7H10O.
First calculate the Degree of Unsaturation:
2C + 2 - (H + X) + N
U=
2
U = Degree of Unsaturation (total number of rings plus pi bonds)
C = number of carbons
H = number of hydrogens
X = number of halogens (F, Cl, Br, I)
N = number of nitrogens
Note: The number of oxygen atoms does not appear in this equation.
2C + 2 - (H + X) + N
2(7) + 2 - (10 + 0) + 0
U=
=
2
2
14 + 2 - (10 ) + 0
U
16 - 10
=
=
2
2
6
= 3
=
2
Note: Be careful with the order of operations in this equation.
Many people get the wrong answer.
Second, draw a structure that has 7 carbons and a degree of unsaturation of 3 (total
number of rings and pi bonds). Include the oxygen atom in this structure, keeping in
mind that oxygen gets two bonds. If you do this, you will automatically have the
right number of hydrogen atoms.
O
OH
C7H10O
C7H10O
O
C7H10O
Problems:
2. Calculate the Degree of Unsaturation then provide two possible skeletal structures for
each molecular formula. Draw in lone pairs of electrons where appropriate.
a. C4H9N
b. C5H10O2
c. C3H7NO
d. C6H10Cl2
e. C11H11NO
C: Formal Charges
Formal charges are very important and must be included in our drawings. If you do
not draw a formal charge when it is supposed to be drawn, then the drawing is
incomplete (and wrong). When calculating formal charges, you need to first consider
the number of valence electrons the atom is supposed to have (get this number from
the periodic table). Second, look at the drawing and ask yourself how many electrons
the atom actually possesses. Each atom owns its lone pair electrons and half of all the
bonding electrons. Finally, determine the difference between these two numbers and
see if you have an excess of electrons (negative charge), a deficiency of electrons
(positive charge) or exactly the right number.
Examples:
Determine the formal charges for the following compounds.
O
a.
b.
c.
N
O
Cl
Br
O
d.
e.
f.
N
Answers:
nc
O
a.
b.
Br
d.
N
c.
O
Cl
nc
O
e.
f.
N
Problems:
3. For each of the compounds below determine if the oxygen or nitrogen atom in the
molecule has a formal charge. If there is a charge, draw the charge near the atom and
circle it. If the there is no formal charge, write “nc” near the molecule.
O
O
N
a.
b.
c.
N
N
e.
f
N
d.
4. In the next exercise you will encounter some atoms with a formal charge. Remember
that carbon is in group 4 of the periodic table and is therefore usually seen with 4
bonds. If the carbon has a (+) charge, then it is because carbon only possesses 3
electrons so that means that there is one electron (and therefore one bond) missing. If
the carbon has a (-) charge, then carbon must own 5 electrons which means it will
have three bonds and one lone pair of electrons. A minus charge automatically
signifies an extra lone pair of electrons. For each of the following problems, take a
look at where the charges are located and draw in all missing lone pairs of electrons.
The lone pairs of electrons will be on certain carbon, nitrogen and oxygen atoms.
N
O
O
O
NH2
a.
b.
c.
d.
D: Resonance Structures
In the next two questions, we will tackle resonance structures. A combination of
resonance structures more accurately describes a particular molecule and helps us better
understand the electronic nature of the compound of interest. Instead of just guessing
what a resonance structure will look like, it is helpful to draw the arrows leading from
one structure to the next. There are a few things you should keep in mind. There are two
important “commandments” that you can never violate when pushing arrows.
Example
a. Thou shall not break a single bond.
b. Thou shall not violate the octet rule.
N
N
Problems:
5. Draw the curved arrows that get you from one resonance drawing to the next. When
you draw arrows, you must begin at the source of electrons such as lone pairs and
pi bonds. If you are using pi bond electrons, begin your arrow halfway across the
bond.
O
O
O
O
a.
b.
OH
O
O
N
N
O
O
OH
d.
c.
6. In the following resonance structure problems, draw the resulting resonance structure
after pushing the arrows shown. Be sure to include formal charges. Draw in all lone
pairs of electrons.
O
N
O
b.
a.
O
c.
O
O
d.